Answer:
[tex]\frac{3}{4}(m+n)-\frac{1}{4}(m-n) = \frac{14}{24}[/tex]
Step-by-step explanation:
Given
[tex]\frac{3}{4}(m+n)-\frac{1}{4}(m-n);m=\frac{1}{2},n=\frac{1}{3}[/tex]
Required
Solve
To do this, we simply substitute values of m and n in the given expression
[tex]\frac{3}{4}(\frac{1}{2}+\frac{1}{3})-\frac{1}{4}(\frac{1}{2}-\frac{1}{3})[/tex]
Evaluate the brackets
[tex]\frac{3}{4}(\frac{3 + 2}{6})-\frac{1}{4}(\frac{3-2}{6})[/tex]
[tex]\frac{3}{4}(\frac{5}{6})-\frac{1}{4}(\frac{1}{6})[/tex]
Open Brackets
[tex]\frac{15}{24} - \frac{1}{24}[/tex]
Take LCM
[tex]\frac{15 - 1}{24}[/tex]
[tex]\frac{14}{24}[/tex]
Hence:
[tex]\frac{3}{4}(m+n)-\frac{1}{4}(m-n) = \frac{14}{24}[/tex]
